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\title{第六章：参数估计}
\author{MSS ET AL}
\date{2018年4月}

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\begin{frame}{本章知识点}

\begin{itemize}

\item  参数的点估计，无偏性，有效性，相合性

\item  似然函数，最大似然估计，EM算法

%一致最小均方误差估计，
\item  一致最小方差无偏估计，充分性，CR不等式

\item  统计推断，贝叶斯公式，贝叶斯估计

\item  区间估计，枢轴量法，置信区间

\item  正态总体参数估计，其它总体参数估计

\end{itemize}

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\begin{frame}{本章习题}

\begin{itemize}

\item  (6.1) 1, 2, 4, 5, 6, 9, 11.

\item  (6.2) 1, 2, 3, 5, 6 ,7.

\item  (6.3) 1, 4, 6, 7, 10.

\item  (6.4) 1, 2, 4, 5, 6, 7.

\item  (6.5) 1, 2, 3, 11, 12.

\item  (6.6) 1, 4, 5, 6, 8, 9, 10, 12, 14, 17.


\end{itemize}


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问：总体为什么会有{\color{red}参数}？为什么要估计参数？


\begin{itemize}
\item 设总体是1万件产品，每件产品的合格率 $p$ 是一个参数。
\item 很难检查全部产品，只好抽查部分产品，然后用这部分产品的情况来估计总体的合格率 $p$.
\item 这里总体服从二点分布 $X\sim b(1,p)$.
\end{itemize}



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问：什么是点估计？


\begin{itemize}
\item 构造一个样本的函数 $\hat{\theta}=\hat{\theta}(X_1,\cdots,X_n)$ 用来估计总体分布里的未知参数 $\theta$, 这个统计量$\hat{\theta}$ 称为 $\theta$ 的点估计。
%\item 参数估计的两种形式：点估计和区间估计。
\item 人们很难完全正确知道总体的参数。{\color{red}点估计}给出一个估计量（估计值）。{\color{red}区间估计}给出估计上限和估计下限。
\end{itemize}



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问：什么是{\color{red}无偏估计}？证明样本均值是总体均值的无偏估计。


\begin{itemize}
\item 若参数的估计量的期望等于该参数，则称该估计量是该参数的无偏估计。
\item 证：$E[\bar{X}]=E[X]=\mu$.
\end{itemize}



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问：设正态总体。证明样本方差是总体方差的无偏估计，但样本标准差不是总体标准差的无偏估计。


\begin{itemize}
\item 样本方差 $S^2=\frac{1}{n-1}\sum_{i=1}^n (X_i-\bar{X})^2 $.
\item 证：$E[S^2]=Var(X)$.
\item 一般地，$E[S]\neq \sqrt{Var(X)}$.
\end{itemize}



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问：设总体服从二点分布 $X\sim b(1,p)$, $0<p<1$. 设样本 $X_1,\cdots, X_n$. 
则参数 $\theta=\frac{1}{p}$ 不存在无偏估计。


\begin{itemize}
\item $T=X_1+\cdots+X_n$ 是充分统计量，$\theta$ 的无偏估计可以写成 $\hat{\theta}=\hat{\theta}(T)$ 的形式。
\item 由无偏估计的定义，$E(\hat{\theta})=\theta$. 这导致矛盾。
%\item 
\end{itemize}



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问：设参数 $\theta$ 有两个无偏估计 $\hat{\theta}_1$ 和 $\hat{\theta}_2$. 什么时候称其中一个比另一个{\color{red}更有效}？


\begin{itemize}
\item 前提是两个都是无偏估计。
\item $Var(\hat{\theta}_1)\le Var(\hat{\theta}_2)$. 
\item 一个估计量的方差越小，则称越有效。
\end{itemize}



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问：设总体分布任意，设总体均值为 $\mu$. \\ 设有样本 $X_1,\cdots,X_n$. 
\begin{itemize}
\item 证明 $\hat{\mu}_1=X_1$ 和 $\hat{\mu}_2=\bar{X}$ 都是 $\mu$ 的无偏估计。
\item 判断哪个更有效。
\end{itemize}


\begin{itemize}
\item $E[X_1]=\mu$, $E[\bar{X}]=\mu$.
\item $Var(X_1)=Var(X)$, $Var(\bar{X})=Var(X)/n$.
\end{itemize}



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问：设总体在 $(0,\theta)$ 均匀分布。设 $X_1,\cdots,X_n$ 是一个样本。顺序统计量为 $X_{(1)}\le\cdots\le X_{(n)}$.  
\begin{itemize}
\item $X_{(n)}$ 是 $\theta$ 的无偏估计吗？
\item 证明 $\hat{\theta}_1:=\frac{n+1}{n}X_{(n)}$ 是 $\theta$ 的一个无偏估计。
\item 证明 $\hat{\theta}_2:=2\bar{X}$ 是 $\theta$ 的一个{\color{red}更有效的无偏估计}。
\end{itemize}


答：直接计算这些估计量的均值和方差。



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问：什么是{\color{red}替换原理}？替换原理的目的是什么？


\begin{itemize}
\item 用样本矩去替换总体矩。
\item 用样本矩的函数去替换总体矩的函数。
\item 目的是用样本来估计总体的参数。
\end{itemize}



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问：设总体是某型号汽车使用五升汽油的行驶路程。测试了20辆这样的汽车，记录他们使用五升汽油的行驶路程，
得到数据为 29.8, 27.6, 28.3, 27.9, 30.1, 28.7, 29.9, 28.0, 27.9, 28.7, 28.4, 27.2, 29.5, 28.5, 28.0, 30.0, 29.1, 29.8, 29.6, 26.9. 求总体均值、总体方差和中位数的{\color{red}矩估计}。


答：计算样本均值、样本方差和样本中位数。
它们就是总体均值、总体方差和中位数的{\color{red}矩估计}。


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\begin{itemize}
\item 问：设总体为指数分布，总体均值为 $1/\lambda$. 设 $X_1,\cdots,X_n$ 是一个样本。求参数 $\lambda$ 的矩估计。
\item 问：设总体为均匀分布 $U(a,b)$, 设 $X_1,\cdots,X_n$ 是一个样本。求参数 $a$ 与 $b$ 的矩估计。
\end{itemize}


答：使用替换原理，用样本矩替换总体矩，然后求解参数。



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问：什么是{\color{red}相合估计量}？


\begin{itemize}
\item 参数 $\theta$ 的估计量 $\hat{\theta}_n$ 满足下述条件，则称 $\hat{\theta}_n$ 是 $\theta$ 的相合估计量：
\[ \lim\limits_{n\to\infty} P(|\hat{\theta}_n-\theta|\ge\varepsilon) =0,\,\, \forall \varepsilon>0. \]
\item 随着样本容量的增加，相合估计量越来越接近参数的真值（概率测度收敛）。
\end{itemize}


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问：设正态总体 $N(\mu,\sigma^2)$, 设 $X_1,\cdots,X_n$ 是一个样本。证明样本均值是总体均值的相合估计，样本方差是总体方差的相合估计。


答：由大数定律估计下述概率得证：
\[ P\left( \Big{|}\bar{X}-\mu \Big{|} \ge \varepsilon \right) \le \frac{Var(\bar{X})}{\varepsilon^2} = \]



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证明：设 $\hat{\theta}_n=\hat{\theta}_n(X_1,\cdots, X_n)$ 是 $\theta$ 的一个估计量，
若下述成立，则 $\hat{\theta}_n$ 是 $\theta$ 的相合估计：
\[ \lim\limits_{n\to\infty} E(\hat{\theta}_n)=\theta, \,\, \lim\limits_{n\to\infty} Var(\hat{\theta}_n)=0. \]


\begin{itemize}
\item 这是{\color{red}相合估计的充分条件}：相合估计量的均值收敛于参数真值，相合估计量的方差收敛于零。
\item 证明是按照定义来{\color{blue}估计概率}。
\end{itemize}



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问：设总体为均匀分布 $U(0,\theta)$, 设 $X_1,\cdots,X_n$ 是一个样本。证明最大值 $X_{(n)}$ 是参数 $\theta$ 的相合估计。


答：直接计算该统计量的均值和方差，证明该统计量符合相合统计量的两个充分条件。

\begin{itemize}
\item $\lim\limits_{n\to\infty} E[X_{(n)}]=\theta.$
\item $\lim\limits_{n\to\infty} Var[X_{(n)}]=0.$
\end{itemize}




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问：设某试验会有三种结果，概率分别为 
\[ p_1=\theta^2,\,\, p_2=2\theta(1-\theta),\,\, p_3=(1-\theta)^2. \]
现在做了 $n$ 次试验，观测到三种结果发生的次数分别为 $n_1$, $n_2$, $n_3$. 试求参数 $\theta$ 的相合估计。


答：由替换原理，用频率替换概率，找到参数的点估计。然后判断这个点估计量是不是相合估计量。



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问：如何理解{\color{red}最大似然原理}？


\begin{itemize}
\item 设有外形完全相同的两个箱子，甲箱中有99个白球和1个黑球，乙箱中有99个黑球和1个白球。现在随机选取一个箱子，然后从中随机选取1球。若发现是白球，问刚才选的是甲箱的概率是多少？
\item 最大似然原理：已经发生的事实是按照最大概率来发生的。
\end{itemize}



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问：考虑产品是否合格。总体为二点分布 $b(1,p)$, 其中 $p$ 是未知的不合格率。设 $X_1,\cdots,X_n$ 是一个样本。写出似然函数，并求 $p$ 的{\color{red}最大似然估计}。


\begin{itemize}
\item 先写出似然函数（定义：就是样本发生的概率），然后求参数使得似然函数达到最大。
\item  $L(p):=P[X_1=x_1,\cdots, X_n=x_n]=$
\end{itemize}






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问：设某试验会有三种结果，概率分别为 
\[ p_1=\theta^2,\,\, p_2=2\theta(1-\theta),\,\, p_3=(1-\theta)^2. \]
现在做了 $n$ 次试验，观测到三种结果发生的次数分别为 $n_1$, $n_2$, $n_3$. 试求参数 $\theta$ 的{\color{red}最大似然估计}。


\begin{itemize}
\item 写出似然函数：$L(\theta)=p_1^{n_1}p_2^{n_2}p_3^{n_3}=$
\item 求 $\theta$ 使 $L(\theta)$ 达到最大。
\end{itemize}




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问：设正态总体 $N(\mu,\sigma^2)$. 设有样本 $X_1,\cdots,X_n$. 求这两个参数的{\color{red}最大似然估计}。


\begin{itemize}
\item 似然函数是个二元函数。求二元函数的最大值，先令这个二元函数的两个偏导数都为零。
\item $L(\mu,\sigma^2)=\Pi p(x_i;\mu,\sigma^2)$ 的最大值在哪里？
\end{itemize}




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问：理解和应用{\color{red}最大似然估计(MLE)}的不变性。


\begin{itemize}
\item 若 $\hat{\theta}$ 是 $\theta$ 的最大似然估计，则 $g(\hat{\theta})$ 是 $g(\theta)$ 的最大似然估计。
\item 应用：求正态总体的0.90分位数的MLE.
%\item 
\end{itemize}




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问：设一次试验可能有四个结果，发生概率分别是 $\frac{2-\theta}{4}$, $\frac{1-\theta}{4}$, $\frac{1+\theta}{4}$, $\frac{\theta}{4}$. 其中 $\theta\in (0,1)$. 现进行 197次试验，四种结果发生的次数分别是 75, 18, 70, 34. 求 $\theta$ 的极大似然估计。
(引进潜变量 $z_1, z_2$)



\begin{table}[ht]\centering
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
概率 & $\frac{1}{4}$ & $\frac{1-\theta}{4}$ & $\frac{1-\theta}{4}$ & $\frac{1}{4}$ & $\frac{\theta}{4}$ & $\frac{\theta}{4}$\\
\hline
次数 & $y_1-z_1$& $z_1$& $y_2$ & $z_2$ & $y_3-z_2$ & $y_4$ \\
\hline
次数 & $75-z_1$& $z_1$& $18$ & $z_2$ & $70-z_2$ & $34$ \\
\hline
\end{tabular}
\end{table}

%将概率分布转化为 $\frac{1}{4}$, $\frac{1-\theta}{4}$, $\frac{1-\theta}{4}$, $\frac{1}{4}$, $\frac{\theta}{4}$, $\frac{\theta}{4}$. 发生次数分别是 $y_1-z_1, z_1, y_2, z_2, y_3-z_2, y_4$. 其中 $z_1,z_2$ 未知。




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问：解释估计参数 $\theta$ 的{\color{red}EM算法}的两个步骤。


\begin{itemize}
\item 写出对数似然函数 $\ell(\theta; y,z)$, 潜变量 $z$ 有分布。
\item 初始：设初始值 $\theta^{(0)}=0.5$ 来估计 $z$ 的分布。
\item E1步：{\color{red}求出期望} $Q(\theta; y,\theta^{(0)}):=\mathbb{E}_z[\ell(\theta;y,z)]$.
\item M1步：{\color{red}极大似然} $\theta^{(1)}:=\underset{\theta}{\mathrm{argmax}}\,\, Q(\theta; y,\theta^{(0)})$.
\item E2步：参数值 $\theta^{(1)}$ 更新 $z$ 的分布，再求期望。
%\item E2步，M2步；如此继续，直到 $\theta^{(n)}$ 收敛。
\end{itemize}




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问：评价点估计的一般标准是什么？


答：设参数 $\theta$ 有点估计量 $\hat{\theta}$, 则下述{\color{red}均方误差}越小，该点估计量越好：
\begin{eqnarray*}
 \mathrm{MSE}_{\theta} (\hat{\theta}) 
 &=& \mathbb{E} [ (\hat{\theta}-\theta)^2 ] \\
&=& \mathrm{Var}_{\theta}(\hat{\theta}) + (\mathbb{E}\hat{\theta}-\theta)^2
\end{eqnarray*}
均方误差 = {\color{red}估计量的方差} + {\color{red}估计量的偏差}的平方



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问：什么是一致最小均方误差估计？


答：设总体分布有参数 $\theta$, 参数空间是 $\Theta$. 称 $\hat{\theta}$ 是{\color{red}一致最小均方误差估计}，如果 $\hat{\theta}$ 是使得均方误差最小的那个估计量，即对其它估计量 $\tilde{\theta}$, 都有：
\[ \mathrm{MSE}_{\theta}(\hat{\theta}) \le \mathrm{MSE}_{\theta}(\tilde{\theta}),\,\, \forall \theta\in\Theta. \]



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问：什么是一致最小方差无偏估计？


答：设总体分布有参数 $\theta$, 设参数空间是 $\Theta$. 称 $\hat{\theta}$ 是{\color{red}一致最小方差无偏估计(UMVUE)}，如果对其它无偏估计 $\tilde{\theta}$，都有
\[ \mathrm{Var}_{\theta}(\hat{\theta}) \le \mathrm{Var}_{\theta}(\tilde{\theta}),\,\, \forall \theta\in\Theta. \]



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问：写出{\color{red}UMVUE}的一个判别准则。


答：设总体分布有参数 $\theta$, 设 $X_1,\cdots, X_n$ 是一个样本。设 $\hat{\theta} = \hat{\theta}(X_1,\cdots,X_n)$ 是一个无偏估计，且 $\mathrm{Var}(\hat{\theta})<\infty$. 则 $\hat{\theta}$ 是 $\theta$ 的{\color{red}一致最小方差无偏估计}的充要条件是：
对任意均值为零、方差有限的统计量 $\varphi=\varphi(X_1,\cdots,X_n)$, 
%若 $\mathbb{E}(\varphi)=0$, $\mathrm{Var}(\varphi)<\infty$, 则
都有 $\mathrm{Cov}_{\theta}(\hat{\theta},\varphi)=0$, $\forall \theta\in\Theta$.




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设总体 $X$ 服从指数分布，总体均值为 $\theta$. 设样本 $X_1,\cdots,X_n$. 记 $T=X_1+\cdots+X_n$, $\bar{X}=T/n$. 证明：
\begin{itemize}
\item $T$ 是 $\theta$ 的充分统计量，$\bar{X}$ 是 $\theta$ 的无偏估计。
\item $\bar{X}$ 是 $\theta$ 的{\color{red}一致最小方差无偏估计}。
\end{itemize}


答：对任意均值为零的统计量 $\varphi$, 按照定义计算协方差 $\mathrm{Cov}_{\theta}(\bar{X},\varphi)=\cdots =0$.




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问：解释统计推断中的{\color{red}充分性原则}。


\begin{itemize}
\item 如果无偏估计不是充分统计量的函数，那么将该估计对充分统计量求条件期望，可以得到一个新的无偏估计，该无偏估计的方差会变小。
\item 设 $T$ 是充分统计量，$\hat{\theta}$ 是一个无偏估计，则 $\tilde{\theta}:=E(\hat{\theta|T})$ 是一个方差更小的无偏估计。证明：直接计算 $\mathrm{Var}(\hat{\theta}) = \mathrm{Var}(\tilde{\theta}) + \mathbb{E}[(\hat{\theta}-\tilde{\theta})^2]$.
\end{itemize}

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{\color{red}充分性原则的例子}：设总体是二点分布 $b(1,p)$, 设 $X_1,\cdots,X_n$ 是样本。设 $T=X_1+\cdots+X_n$. 设 $\theta=p^2$. 作统计量 $\hat{\theta}_1=1_{x_1}1_{x_2}$. 令 $\hat{\theta}:=\mathbb{E}(\hat{\theta}_1\, |\, T=t)$. 则有：
\begin{itemize}
\item $\hat{\theta}_1$ 是 $\theta$ 的无偏估计。
\item $T$ 是充分统计量。
\item $\hat{\theta}$ 也是 $\theta$ 的无偏估计，且方差更小。
\end{itemize}





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问：什么是费希尔信息量？


答：设总体的概率函数是 $p(x;\theta), \theta\in\Theta$ 满足某些条件(i,ii,iii,iv,v)。称下述 $\theta$ 的函数表达式为总体分布的{\color{red}费希尔信息量}：
\[I(\theta) = \mathbb{E}\left[ \frac{\partial}{\partial \theta} \ln p(x;\theta) \right]^2.\] 




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问：计算泊松分布和指数分布的费希尔信息量。


\begin{itemize}
\item $X\sim P(\lambda)$,\,\,\, $I(\lambda)=1/\lambda$.
\item $X\sim Exp(1/\theta)$,\,\,\, $I(\theta)=1/\theta^2$.
\end{itemize}



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问：解释{\color{red}Cramer-Rao不等式}。


\begin{itemize}
\item 设 $T$ 是 $g(\theta)$ 的无偏估计。函数 $g$ 符合条件。
\item CR不等式：$\mathrm{Var}(T)\ge g'(\theta)^2/(nI(\theta))$.
\item CR不等式给出了无偏估计的方差的下界。
\item 达到该下界的无偏估计是UMVUE.
\end{itemize}




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{\color{red}充分性原则的例子继续}：设总体分布 $b(1,\theta)$. 
\begin{itemize}
\item 计算该总体分布的费希尔信息量。
\item 求参数 $\theta$ 的CR下界。
\item 证明样本均值 $\bar{X}$ 是 $\theta$ 的UMVUE.
\end{itemize}


答：这里 $g(\theta)=\theta$. 因 $\mathrm{Var}(\bar{X})=1/(nI(\theta))$, 由CR不等式得证。


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问：设总体是指数分布 $Exp(1/\theta)$. 
\begin{itemize}
\item 计算该总体分布的费希尔信息量。
\item 求参数 $\theta$ 的CR下界。
\item 证明样本均值 $\bar{X}$ 是 $\theta$ 的UMVUE.
\end{itemize}


答：也是要验证 $\mathrm{Var}(\bar{X})=1/(nI(\theta))$.



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问：贝叶斯学派的统计推断的基础是什么？


\begin{itemize}
\item 总体信息：总体分布族提供的信息。
\item 样本信息：抽取样本所得观测值提供的信息。
\item 先验信息：抽样前已知，来源经验历史资料。
\end{itemize}



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问：什么是参数 $\theta$ 的{\color{red}先验分布}和{\color{red}后验分布}？


\begin{itemize}
\item 贝叶斯学派认为参数 $\theta$ 也是随机变量。
\item 从先验信息得到的参数 $\theta$ 的分布称先验分布。
\item 参数 $\theta$ 的{\color{red}先验分布}经过{\color{red}总体信息}和{\color{red}样本信息}的统计推断之后形成的新的分布，称为参数 $\theta$ 的{\color{red}后验分布}。
\end{itemize}


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问：解释{\color{red}贝叶斯公式}的密度函数形式。


\begin{itemize}
\item 公式： $\pi(\theta\, |\, x) = \frac{h(x,\theta)}{m(x)} 
= \frac{p(x\,|\,\theta)\pi(\theta)}{\int_\Theta p(x\,|\,\theta)\pi(\theta)d\theta  } $.
\item 先验分布 $\pi(\theta)$, 后验分布 $\pi(\theta\,|\, x)$.
\item 样本的联合概率函数 $p(x\,|\,\theta)=\Pi_{i=1}^{n} p(x_i\,|\,\theta)$, \\ 综合了总体信息和样本信息。
\item 样本观测值 $x=(x_1,\cdots,x_n)$.
\end{itemize}




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问：如何从后验分布 $\pi(\theta\,|\, x)$ 得出参数的估计？


\begin{itemize}
\item 最大后验估计：使用后验分布的最大值。
\item 后验中位数估计：使用后验分布的中位数。
\item 后验期望估计：使用后验分布的均值。
\end{itemize}



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问：设某事件 A 在一次试验中发生的概率为 $\theta$. 设先验信息为无，为估计 $\theta$ 对试验进行了$n$ 次独立观测，其中事件 A 发生了 $x$ 次。写出先验分布，样本和参数的联合分布，并求参数的后验期望估计。


\begin{itemize}
\item 参数的先验分布 $\theta\sim U(0,1)$.
\item 联合分布 $h(x,\theta)=\binom{n}{x}\theta^x(1-\theta)^{n-x}, (x,\theta)\in\square$.
\item 后验期望估计 $\hat{\theta}_B = E(\theta\,|\,x)=\frac{x+1}{n+2}$.
\end{itemize}




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问：设 $X_1,\cdots,X_n$ 是来自正态分布 $N(\mu,\sigma_0^2)$ 的一个样本，其中 $\sigma_0^2$ 已知，$\mu$ 未知，假设 $\mu$ 的先验分布也是正态分布 $N(\theta,\tau^2)$, 其中先验均值 $\theta$ 和先验方差 $\tau^2$ 均已知，求 $\mu$ 的贝叶斯估计。


\begin{itemize}
\item 贝叶斯公式 $\pi(\mu\,|\,x) = \frac{h(x,\mu)}{m(x)}$.
\item 贝叶斯估计 $\hat{\mu}=\frac{n/\sigma_0^2}{n/\sigma_0^2+1/\tau^2} \bar{x} + \frac{1/\tau^2}{n/\sigma_0^2+1/\tau^2}\theta$.
\end{itemize}


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问：总体参数 $\theta$ 的{\color{red}区间估计}是什么？


\begin{itemize}
\item 是区间 $\hat{\theta}_L\le \theta\le \hat{\theta}_U$, 两个端点都是统计量。
\item $\hat{\theta}_L=\hat{\theta}_L(X_1,\cdots,X_n)$. L = Lower.
\item $\hat{\theta}_U=\hat{\theta}_U(X_1,\cdots,X_n)$. U = Upper.
\item 对任意 $\theta\in \Theta$, 有 $P[\hat{\theta}_L\le \theta\le \hat{\theta}_U]\ge 1-\alpha$.
\item 称 $1-\alpha$ 为置信水平。
\end{itemize}


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问：设总体是正态分布 $N(\mu,\sigma^2)$. 设 $X_1,\cdots,X_n$ 是样本。 
求 $\mu$ 的置信水平为 $1-\alpha$ 的置信区间。


答：由``数理统计基本定理''，统计量 $\frac{\bar{X}-\mu}{s/\sqrt{n}}$ 服从 $t(n-1)$ 分布，
因此得到下式，并解得置信区间。
\[ P\left[ -t_{1-\frac{\alpha}{2}}(n-1) \le \frac{\bar{X}-\mu}{s/\sqrt{n}} \le t_{1-\frac{\alpha}{2}}(n-1)  \right] = 1-\alpha \]




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问：如何理解单侧置信区间？


\begin{itemize}
\item 设参数是产品寿命，只想知道其置信下限。
\item 对任意 $\theta\in \Theta$, 有 $P[\hat{\theta}_L\le \theta]\ge 1-\alpha$.
\item $\hat{\theta}_L=\hat{\theta}_L(X_1,\cdots,X_n)$.
\end{itemize}


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问：构造参数的置信区间的常用方法是{\color{red}枢轴量法}，其步骤是什么？


\begin{itemize}
\item 构造枢轴量 $G=G(X_1,\cdots,X_n,\theta)$.
\item 选常数 $c,d$ 使得 $P[c\le G\le d]=1-\alpha$.
\item 解出 $\theta$ 的范围。
\end{itemize}



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问：设总体是均匀分布 $U(0,\theta)$. 设 $X_1,\cdots,X_n$ 是样本。求参数 $\theta$ 的最短置信区间。


\begin{itemize}

\item 选取枢轴量 $G=X_{(n)}/\theta$. 其中 $X_{n}$ 是最大次序统计量。(怎么想到的？)
\item 写出 $G$ 的分布函数，与参数 $\theta$ 无关。
\item 由 $P[c\le G\le d] = d^n-c^n$ 解得置信区间。
\end{itemize}





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问：用天平秤量某物体的质量9次，得平均值 $\bar{x}=15.4$(g). 设天平秤量结果为正态分布，已知标准差为 $0.1$(g). 求该物体质量的 0.95 置信区间。



\begin{itemize}
\item 选取统计量 $G=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$, 其分布为 $N(0,1)$.
\item 确定分位数 $P(-u_{0.975}<G<u_{0.975})=0.95$.
\item 代入数据解不等式 $-u_{0.975}<\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}<u_{0.975}$.
\end{itemize}


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问：总体参数的区间估计的几点说明。


\begin{itemize}
\item 统计量是随机变量，其分布规律由统计学家们研究得来。常用的是标准正态分布 $N(0,1)$ 和三大抽样分布 $t(n)$, $\chi^2(n)$, $F(m,n)$. 
\item 分布的分位数由查表或统计软件算出。
\item 统计量的密度函数的图像要清楚。
\item 注意大小写的差别，$\bar{x}$ 是 $\bar{X}$ 的观测值。
\end{itemize}



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问：设总体为正态分布 $N(\mu,1)$, 为得到 $\mu$ 的置信水平为 $0.95$ 的置信区间的长度不超过 $1.2$, 样本容量最小多少？


\begin{itemize}
\item 选取统计量 $G=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$.
%\item 确定分位数 $P(-u_{0.975}<G<u_{0.975})=0.95$.
\item 在 $-u_{0.975}<\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}<u_{0.975}$ 中 $\sigma$ 已知。
\item 由 $\mu$ 的置信区间长度为 $1.2$ 解出 $n$.
\end{itemize}



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问：设轮胎寿命服从正态分布，现随机抽取12只轮胎，测得寿命(单位：万千米)如下：4.68, 4.85, 4.32, 4.85, 4.61, 5.02, 5.20, 4.60, 4.58, 4.72, 4.38, 4.70. 求平均寿命的 0.95 置信区间。



\begin{itemize}
\item 选取统计量 $T=\frac{\bar{X}-\mu}{s/\sqrt{n}}$, 其分布为 $t(11)$.
\item 算概率 $P(-t_{0.975}(11)<T<t_{0.975}(11))=0.95$.
\item 代入数据解出 $\mu$ 的置信区间。
\end{itemize}



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问：如果要计算轮胎寿命的置信下限，即以0.95 置信度可以说寿命至少是多少？


\begin{itemize}
\item 仍选取统计量 $T=\frac{\bar{X}-\mu}{s/\sqrt{n}}$, 其分布为 $t(11)$.
\item 算概率 $P(T<t_{0.975}(11))=0.95$.
\item 代入数据解出 $\mu$ 的单侧置信区间。
\item 选哪边的不等式，要看单侧区间的不等式。
\end{itemize}




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问：某厂生产的零件的质量为正态分布 $N(\mu,\sigma^2)$. 现抽取9个样品，测得其质量(单位：g)为
 45.3, 45.4, 45.1, 45.3, 45.5, 45.7, 45.4, 45.3, 45.6. 求总体标准差 $\sigma$ 的 0.95 置信区间。


\begin{itemize}
\item 选取统计量 $\chi^2 = \frac{(n-1)S^2}{\sigma^2}$, 其分布为 $\chi^2(8)$.
\item 算概率 $P(\chi^2_{0.025}(8) < \chi^2 < \chi^2_{0.975}(8) ) =0.95$.
\item 解不等式 $\chi^2_{0.025}(8) <  \frac{(n-1)s^2}{\sigma^2} < \chi^2_{0.975}(8)$.
\end{itemize}


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问：对某事件 A 观察 120 次，$A$ 发生了 36次。求事件 $A$ 的发生概率 $p$ 的置信区间。


\begin{itemize}
\item 总体是二点分布 $X\sim b(1,p)$, 简单随机样本 $X_1,\cdots, X_n$, 样本容量 $n=120$ 较大。
\item 记 $Y=X_1+\cdots+X_n$, 由中心极限定理知，统计量 $U:=\frac{Y-E(Y)}{\sqrt{Var(Y)}}$ 近似于标准正态分布。
\item 计算知 $E(Y)=np$, $Var(Y)=np(1-p)$.
\end{itemize}



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问：为调查某电视节目的收视率 $p$, 为了使其置信区间的长度不超过 $2d_0$, 应至少调查的是用户？


\begin{itemize}
\item 总体是二点分布 $X\sim b(1,p)$, $X_i$ 是第 $i$ 个用户对应的随机变量，取值为1的概率是 $p$.
\item 记 $Y=\sum\limits_{i=1}^{n}X_i$, 作统计量 $U:=\frac{Y-E(Y)}{\sqrt{Var(Y)}}$.
\item 写出概率表达式，写出不等式。
\end{itemize}



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问：为比较两个小麦品种的产量，播种试验田甲品种8块乙品种10块。测得单位面积产量(单位：千克)为：
甲：628, 583, 510, 554, 612, 523, 530, 615; 
乙：535, 433, 398, 470, 567, 480, 498, 560, 503, 426.
设单位面积产量服从正态分布，方差相等。求甲乙单位面积产量差的置信区间 $(\alpha=0.05)$.



答：使用统计量 $T=\sqrt{\frac{mn}{m+n}}\frac{(\bar{X}-\bar{Y})-(\mu_1-\mu_2)}{S_\omega}$. 
其中 $S_\omega^2:=\frac{(m-1)S_1^2+(n-1)S_2^2}{m+n-2}$. 
该统计量服从 $t(m+n-2)$.




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问：设自动机床加工的套筒直径服从正态分布，在两个班次中分别抽取5个和6个，测得直径(单位：cm)为：
甲：5.06, 5.08, 5.03, 5.00, 5.07; 
乙：4.98, 5.03, 4.97, 4.99, 5.02, 4.95.
求两个班次加工的套筒直径的方差比 $\sigma_1^2/\sigma_2^2$ 的 0.95 置信区间。


答：使用统计量 $F=\frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2}$, 其分布为 $F(4,5)$. 
概率表达式为 $P[F_{0.025}(4,5) < F< F_{0.975}(4,5) ] = 0.95$. 
代入 $s_1^2, s_2^2$ 和分位数的值求得置信区间。


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